Question

Let $E=x^3(x^3+1)(x^3+2)(x^3+3);xϵR$. Then minimum value of $E$ be

• A. 1
• B. -2
• C. -1
• D. Minimum value is not attained

Answer 1

The correct option is C -1

$f\left(x\right)=\left(x^3(x^3+1)(x^3+2)(x^3+3)\right)f^{{}^{\prime }}\left(x\right)=6x^2(2x^9+9x^6+11x^3+3)$
Now on solving we get $f^{{}^{\prime }}\left(x\right)=0$ is at $x=0$
At $x=0f\left(x\right)=0$
At $x=\sqrt[3]{3-\frac{3}{2}},x=-{(-1)}^{\frac{2}{3}}\sqrt[3]{3\frac{3}{2}}$and $x=-\sqrt[3]{3\frac{3}{2}}f\left(x\right)=\frac{9}{16}$
At $x=-\frac{\sqrt[3]{33-\sqrt{5}}}{\sqrt[3]{32}},x=\frac{\sqrt[3]{3\sqrt{5}-3}}{\sqrt[3]{32}},x=-\frac{{(-1)}^{\frac{2}{3}}\sqrt[3]{33-\sqrt{5}}}{\sqrt[3]{32}},$

$x=\sqrt[3]{3\frac{1}{2}(-3-\sqrt{5})},x=-\sqrt[3]{3\frac{1}{2}(3+\sqrt{5})}$ and $x=-\left({(-1)}^{\frac{2}{3}}\sqrt[3]{3\frac{1}{2}(3+\sqrt{5})}\right)$

$f\left(x\right)=0$
So minimum value is $-1$ at $x=0$